Someone help: 1 Asalesman leaves home at 0800 and travels for 13 hours at an average
speed of 60km/h. He then stops for 30 minutes. He continues for 2 more
hours at 50km/h and stops for 1 hour. He then returns home and arrives
home at 1600.

a) Show this in a travel graph.
b) Calculate his average speed for his return journey.

a) Please find attached the required travel graph with times increasing from (0:00) of the day of travel to the following day

b) The average speed of return journey = 880 km/(15.5 h) ≈ 56.77 km/h

Step-by-step explanation:

The given information includes;

The time of departure of the salesman, t₁ = 0800

The duration of travel of the salesman = 13 hours

The average speed with which the salesman travels = 60 km/h

The time duration in which he stops = 30 minutes

The speed with which he continues = 50 km/h

The time duration he travels at 50 km/h = 2 hours

The time he arrives home after he turns, t₂ = 16:00

The appropriate graph is a distance time graph

So as to show the given information in a travel graph, we calculate the distances as follows;

Time                                                                       Location

08:00                                                                     0

08:00 + 13 = 21:00                                                  13 × 60 = 780

21:00 + 0.30 = 21:30                                               780

21:30 + 2:00 = 23:30                                              780 + 50 × 2 = 880

23:30 + 1:00 = 0:30 (24: hours )                            880

16:00  (40 hours from 0:00 the previous day)  

Therefore, we have;

Location; 0, 780, 780, 880, 880, 0

Time; 08:00, 21:00, 21:30, 23:30, 24:30, 40:00

To show the distance traveled over the given time periods graphically, we let 00:30 = 24.30 and the arrival time (the next day) 16:00 = 40.00 hours after midnight (0:00) the previous day                

Please find attached the required travel graph created with Microsoft Excel

b) The average speed of return journey = (Total distance of return journey)/(Total time)

The total distance going = 880 km

Therefore the return journey is also 880 km

∴ The total time taken on return journey Δt = (Time at arrival home) - (Time of start of return journey)

∴ Δt = 40 - 24.5 = 15.5

Δt = 15.5 hour

∴ The average speed of return journey = 880 km/(15.5 h) ≈ 56.77 km/h.  

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)